A thin but rigid semicircular wire frame of r | vedclass

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Question

(A) Let $	heta$ be the angle that the radius to the peg $P$ makes with the horizontal. The horizontal distance $x$ of the peg from the hinge $O$ is g

Vedclass · Accepted Answer

$v_0 / r$

Vedclass · Answer

(A) Let $	heta$ be the angle that the radius to the peg $P$ makes with the horizontal. The horizontal distance $x$ of the peg from the hinge $O$ is given by $x = 2r \cos 	heta$. However,based on the geometry of the semicircular frame,the distance from $O$ to the peg $P$ along the horizontal is $x = 2r \sin 	heta$,where $	heta$ is the angle the diameter makes with the vertical.Given that the peg moves horizontally with constant speed $v_0$,we have $v_0 = \frac{dx}{dt}$.Differentiating $x = 2r \sin 	heta$ with respect to time $t$,we get:$\frac{dx}{dt} = 2r \cos 	heta \cdot \frac{d	heta}{dt}$Since $\omega = \frac{d	heta}{dt}$,we have:$v_0 = 2r \cos 	heta \cdot \omega$$\omega = \frac{v_0}{2r \cos 	heta}$Given $	heta = 60^{\circ}$,we have $\cos 60^{\circ} = 0.5$.Substituting the values:$\omega = \frac{v_0}{2r \cdot 0.5} = \frac{v_0}{r}$.

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